| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Perpendicular velocity directions |
| Difficulty | Challenging +1.2 This is a multi-part projectiles question requiring trajectory derivation, range calculation, energy at maximum height, and perpendicular velocity analysis. Parts (a)-(c) are standard M2 fare, but part (d) requires recognizing that perpendicular velocities means the velocity vector has rotated 90°, demanding dot product reasoning or velocity component analysis beyond routine exercises. The multiple steps and the conceptual leap in part (d) place this moderately above average difficulty. |
| Spec | 3.02i Projectile motion: constant acceleration model6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
7. A particle, of mass 0.3 kg , is projected from a point $O$ on horizontal ground with speed $u$. The particle is projected at an angle $\alpha$ above the horizontal, where $\tan \alpha = 2$, and moves freely under gravity. When the particle has moved a horizontal distance $x$ from $O$, its height above the ground is $y$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$y = 2 x - \frac { 5 g } { 2 u ^ { 2 } } x ^ { 2 }$$
The particle hits the ground at the point $A$, where $O A = 36 \mathrm {~m}$.
\item Find $u$, the speed of projection.
\item Find the minimum kinetic energy of the particle as it moves between $O$ and $A$.
The point $B$ lies on the path of the particle. The direction of motion of the particle at $B$ is perpendicular to the initial direction of motion of the particle.
\item Find the horizontal distance between $O$ and $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q7 [15]}}